7.3 Probability of Mutually Exclusive Events

Posted on April 23, 2020 by user

7.3 Probability of Mutually Exclusive Events

1. Two events are mutually exclusive if they cannot occur at the same time.

2. If A and B are mutually exclusive events, then

P (A υ B) = P (A) + P (B)

Example:

A bag contains 3 blue cards, 4 green cards and 5 yellow cards. A card is chosen at random from the box. Find the probability that the chosen card is green or yellow.

Solution:

Let G = event when a green card is chosen.

Y = event when a yellow card is chosen.

The sample space, S = 12, n (S) = 12

n (G) = 4 and n (Y) = 5

\begin{array}{l} P (G) = \frac{n (G)}{n (S)} = \frac{4}{12} \\ P (Y) = \frac{n (Y)}{n (S)} = \frac{5}{12} \end{array}

Events G and Y cannot occur simultaneously because we cannot obtain green card and yellow card at the same time. Therefore, events G and Y are mutually exclusive.

\begin{array}{l} G \cap Y = \emptyset \\ P (G \cup Y) = P (G) + P (Y) \\ = \frac{4}{12} + \frac{5}{12} \\ = \frac{9}{12} = \frac{3}{4} \end{array}

7.4 Probability of Independent Events

Posted on April 23, 2020 by user

7.4 Probability of Independent Events

1. In an experiment, if the outcomes of event A do not influence the outcomes of event B, then the two events are independent.

2. If A and B are two independent events, the probability for the occurrence of events A and B is

P (A ∩ B) = P (A) × P (B)

3. The concept of the probability of two independent events can be expanded to three or more independent events. If A, B and C are three independent events, the probability for the occurrence of events A, B and C is

P (A ∩ B ∩ C) = P (A) x P (B) x P (C)

4. A tree diagram can be constructed to show all the possible outcomes of an experiment.

Example:

Fatimah, Emily and Rani are to take a written driving test. The probability that they pass the test are ½, ⅔ and ¾ respectively. Calculate the probability that

(a) only one of them passes the exam,

(b) at least two of them pass the exam,

Solution:

Let P = Pass and F = Fail

The tree diagram is as follows.

(a)

P (only one of them passes the exam)

= P(PFF or FPF or FFP)

= P(PFF) + P(FPF) + P(FFP)

\begin{array}{l} = \frac{1}{24} + \frac{1}{12} + \frac{1}{8} \\ = \frac{1}{4} \end{array}

(b)

P (at least two of them pass the exam)

= P (PPP or PPF or PFP or FPP)

= P (PPP) + P (PPF) + P (PFP) + P (FPP)

\begin{array}{l} = \frac{1}{4} + \frac{1}{12} + \frac{1}{8} + \frac{1}{4} \\ = \frac{17}{24} \end{array}

(c)

P (at least one of them passes the exam)

= 1 – P (all of them fail)

= 1 – P (FFF)

\begin{array}{l} = 1 - \frac{1}{24} \\ = \frac{23}{24} \end{array}

Long Questions (Question 1)

Posted on April 23, 2020 by user

Question 1:

Peter, William and Roger compete with each other in shooting a target. The probabilities that they strike the target are

\frac{2}{5}, \frac{3}{4} and \frac{2}{3}

respectively. Calculate the probability that

(a) all the three of them strike the target,

(b) only one of them strikes the target,

Solution:

Let S = Strike and M = Missed

\begin{array}{l} Probability of Peter missed the target = \frac{3}{5} \\ Probability of William missed the target = \frac{1}{4} \\ Probability of Roger missed the target = \frac{1}{3} \end{array}

(a)

\begin{array}{l} Probability (all three persons strike the target) \\ = \frac{2}{5} \times \frac{3}{4} \times \frac{2}{3} \\ = \frac{1}{5} \end{array}

(b)

\begin{array}{l} Probability (only one of them strikes the target) \\ = P (only Peter struck) + P (only William struck) + P (only Roger struck) \\ = (\frac{2}{5} \times \frac{1}{4} \times \frac{1}{3}) + (\frac{3}{5} \times \frac{3}{4} \times \frac{1}{3}) + (\frac{3}{5} \times \frac{1}{4} \times \frac{2}{3}) \\ = \frac{1}{30} + \frac{3}{20} + \frac{1}{10} \\ = \frac{17}{60} \end{array}

(c)

\begin{array}{l} Probability (at least one of them strikes the target) \\ = 1 - P (all missed the target) \\ = 1 - (\frac{3}{5} \times \frac{1}{4} \times \frac{1}{3}) \\ = 1 - \frac{1}{20} \\ = \frac{19}{20} \end{array}

7.2 Probability of the Combination of Two Events

Posted on April 23, 2020 by user

7.2 Probability of the Combination of Two Events

1. For two events, A and B, in a sample space S, the events A ∩ B (A and B) and A υ B (A or B) are known as combined events.

2. The probability of the union of sets A and B is given by:

P (A \cup B) = P (A) + P (B) - P (A \cap B)

3. The probability of the union of sets A and B can also be calculated using an alternative method, i.e.

P (A \cup B) = \frac{n (A \cup B)}{n (S)}

4. The probability of event A and event B occurring, P(A ∩ B) can be determined by the following formula.

P (A \cap B) = \frac{n (A \cap B)}{n (S)}

Example:

Given a universal set ξ = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}. A number is chosen at random from the set ξ . Find the probability that

(a) an even number is chosen.

(b) an odd number or a prime number is chosen.

Solution:

The sample space, S = ξ

n(S) = 14

(a)

Let A = Event of an even number is chosen

A = {2, 4, 6, 8, 10, 12, 14}

n (A) = 7

\begin{array}{l} P (A) = \frac{n (A)}{n (S)} \\ = \frac{7}{14} = \frac{1}{2} \end{array}

(b)
Let,

B = Event of an odd number is chosen

C = Event of a prime number is chosen

B = {3, 5, 7, 9, 11, 13, 15} and n (B) = 7

C = {2, 3, 5, 7, 11, 13} and n (C) = 6

The event when an odd number or a prime number is chosen is B υ C.

P (B υ C) = P (B) + P (C) – P (B ∩ C)

B ∩ C = {3, 5, 7, 11, 13}, n (B ∩ C) = 5

\begin{array}{l} P (B \cup C) \\ = P (B) + P (C) - P (B \cap C) \\ = \frac{n (B)}{n (S)} + \frac{n (C)}{n (S)} - \frac{n (B \cap C)}{n (S)} \\ = \frac{7}{14} + \frac{6}{14} - \frac{5}{14} \\ = \frac{8}{14} = \frac{4}{7} \\ ∴ The probability of choosing an \\ odd number or a prime number = \frac{4}{7} . \end{array}

Short Questions (Question 1 to 3)

Posted on April 23, 2020 by user

Question 1:

The probability of student P being chosen as a school prefect is

\frac{3}{4}

while the probability of student Q being chosen is

\frac{5}{6}

Find the probability that

(a) both of the students are chosen as the school prefect,

(b) only one student is chosen as a school prefect.

Solution:

(a)

\begin{array}{l} Probability (both of the students are chosen as the school prefect) \\ = \frac{3}{4} \times \frac{5}{6} \\ = \frac{5}{8} \end{array}

(b)

\begin{array}{l} Probability (only one student is chosen as a school prefect) \\ = (\frac{3}{4} \times \frac{1}{6}) + (\frac{1}{4} \times \frac{5}{6}) \\ = \frac{3}{24} + \frac{5}{24} \\ = \frac{1}{3} \end{array}

Question 2:

A bag contains x pink cards and 6 green cards. Two cards are drawn at random from the bag, one after the other, without replacement. Find the value of x if the probability of obtaining two green cards is ⅓.

Solution:

Total cards in the bag = x + 6

P(obtaining 2 green cards) = ⅓

\begin{array}{l} \frac{6}{x + 6} \times \frac{5}{x + 5} = \frac{1}{3} \\ \frac{30}{(x + 6) (x + 5)} = \frac{1}{3} \end{array}

(x + 6) (x + 5) = 90

x² + 11x + 30 = 90

x² + 11x – 60 = 0

(x – 4) (x + 15) = 0

x = 4 or x = –15 (not accepted)

Question 3:
A sample space of an experiment is given by S = {1, 2, 3, … , 21}. Events Q and R are defined as follows:
Q : {3, 6, 9, 12, 15, 18, 21}
R : {1, 3, 5, 15, 21}

Find
(a) P(Q)
(b) P(Q and R)

Solution:
(a)

\begin{array}{l} n (S) = 21, n (Q) = 7 \\ P (Q) = \frac{7}{21} = \frac{1}{3} \end{array}

(b)

\begin{array}{l} Q \cap R = {3, 15, 21}, \\ t h e n n (Q \cap R) = 3 \\ P (Q and R) = P (Q \cap R) \\ = \frac{n (Q \cap R)}{n (S)} \\ = \frac{3}{21} \\ = \frac{1}{7} \end{array}

Long Questions (Question 2 & 3)

Posted on April 23, 2020 by user

Question 2:

A bag contains 4 red cards, 6 blue cards and 5 green cards. A card is drawn at random and its colour is recorded and then it is returned to the bag before the second card is drawn at random. Find the probability that

(a) all the three cards are blue,

(b) there are two blue cards followed by one red card,

(d) all the three cards have the same colour.

Solution:

Let R = red card, B = blue card, G = green card

(a)

\begin{array}{l} Probability (all the three cards are blue) \\ = \frac{6}{15} \times \frac{6}{15} \times \frac{6}{15} \\ = \frac{8}{125} \end{array}

(b)

\begin{array}{l} Probability (two blue cards followed by one red card) \\ = \frac{6}{15} \times \frac{6}{15} \times \frac{4}{15} \\ = \frac{16}{375} \end{array}

(c)

\begin{array}{l} Probability (the sequence of the cards drawn is red, green and blue) \\ = \frac{4}{15} \times \frac{6}{15} \times \frac{5}{15} \\ = \frac{8}{225} \end{array}

(d)

\begin{array}{l} Probability (all the three cards have the same colour) \\ = P (R R R) + P (B B B) + P (G G G) \\ = {(\frac{4}{15})}^{3} + {(\frac{6}{15})}^{3} + {(\frac{5}{15})}^{3} \\ = \frac{64}{3375} + \frac{216}{3375} + \frac{125}{3375} \\ = \frac{3}{25} \end{array}

Question 3:

A bag contains 4 blue beads, 3 red beads and 7 green beads. Two beads are drawn at random from the bag, one after the other without replacement. Find the probability that

(a) both the beads are of the same colour,

(b) both the beads are of different colours.

Solution:

(a)

\begin{array}{l} Probability (both the beads are of the same colour) \\ = P (blue, blue) + P (red, red) + P (green, green) \\ = (\frac{4}{14} \times \frac{3}{13}) + (\frac{3}{14} \times \frac{2}{13}) + (\frac{7}{14} \times \frac{6}{13}) \\ = \frac{6}{91} + \frac{3}{91} + \frac{3}{13} \\ = \frac{30}{91} \end{array}

(b)

\begin{array}{l} Probability (both the beads are of different colour) \\ = 1 - P (both the beads are of the same colour) \\ = 1 - \frac{30}{91} \leftarrow From part (a) \\ = \frac{61}{91} \end{array}

6.1 Permutation Part 1

Posted on April 23, 2020 by user

(A) rs Multiplication Principle/ Rule

1. If an operation can be carried out in r ways and another operation can be carried out in s ways, then the number of ways to carry out both the operations consecutively is r × s, i.e. rs.

2. The rs multiplication principle can be expanded to three or more operations. If the numbers of ways for the occurrence of events A, B and C are r, s and p respectively, the number of ways for the occurrence of all the three events consecutively is r × s × p, i.e. rsp.

Example 1:
There are 3 different roads to travel from town P to town Q and 4 different roads to travel from town Q to town R. Calculate the number of ways a person can travel from town P to town R via town Q.

Solution:

3 × 4 = 12

(B) Permutations

Example 2:
Calculate each of the following.
(a) 7!
(b) 4!6!
(c) 0!5!

\begin{array}{l} (d) \frac{7!}{5!} \\ (e) \frac{8!}{4!} \\ (f) \frac{n!}{(n - 2)!} \\ (g) \frac{n! 0!}{(n - 1)!} \\ (h) \frac{3! (n + 1)!}{2! n!} \end{array}

Solution:

(a) 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

(b) 4!6! = (4 × 3 × 2 × 1)( 6 × 5 × 4 × 3 × 2 × 1) = 17280

(c) 0!5! = (1)( 5 × 4 × 3 × 2 × 1) = 120

\begin{array}{l} (d) \frac{7!}{5!} = \frac{7 \times 6 \times 5!}{5!} = 7 \times 6 = 42 \\ (e) \frac{8!}{4!} = \frac{8 \times 7 \times 6 \times 5 \times 4!}{4!} = 8 \times 7 \times 6 \times 5 = 1680 \\ (f) \frac{n!}{(n - 2)!} = \frac{n (n - 1) (n - 2)}{(n - 2)} = n (n - 1) \\ (g) \frac{n! 0!}{(n - 1)!} = \frac{n (n - 1) (1)}{(n - 1)} = n \\ (h) \frac{3! (n + 1)!}{2! n!} = \frac{3 \times 2! (n + 1) (n) (n - 1)}{2! n (n - 1)} = 3 (n + 1) \end{array}

Calculator Computation:

Short Question 1 – 3

Posted on April 23, 2020 by user

Question 1:
A group of 4 men and 3 ladies are to be seated in a row for a photographing session. If the men and ladies want to be seated alternately (man-lady-man-lady...), calculate the number of different arrangements.

Solution:
The arrangements of 4 men and 3 ladies to be seated alternately are as follow:

M L M L M L M

The number of ways to arrange the seat for 4 men = 4!

The number of ways to arrange the seat for 3 ladies = 3!

Total number of different arrangements for men and ladies = 4! × 3! = 144

Question 2:
Ahmad has 6 durians, 5 watermelons and 2 papayas. If he wants to arrange these fruits in a row and the fruits of the same kind have to be grouped together, calculate the number of different arrangements. The sizes of the fruits are different.

Solution:
The number of ways to arrange 3 groups of fruits that are same kind = 3!

DDDDDD WWWWW PP

The number of ways to arrange 6 durians = 6!
The number of ways to arrange 5 watermelons = 5!
The number of ways to arrange 2 papayas = 2!

Therefore, the number of ways to arrange the fruits with same kind of fruits was grouped together
= 3! × 6! × 5! × 2!
= 1036800

Question 3:
Calculate the number of arrangements, without repetitions, of the letters from the word `SOMETHING' with the condition that they must begin with a vowel.

Solution:

\begin{array}{l} Arrangement of letters begin with vowel \\ = O, M and I \\ =^{3} P_{1} \\ Arrangement for the rest of the letters \\ = 7! \\ ∴ Number of arrangements \\ =^{3} P_{1} \times 7! \\ = 15120 \end{array}

6.3 Combinations

Posted on April 23, 2020 by user

6.3 Combinations

(1) The number of combinations of r objects chosen from n different objects is given by :

^{n} C_{r} = \frac{n!}{r! (n - r)!}

(2) A combination of r objects chosen from n different objects is a selection of a set of r objects chosen from n objects. The order of the objects in the chosen set is not taken into consideration.

\begin{array}{l} N o t e : \\ (i)^{n} C_{0} = 1 \\ (i i)^{n} C_{n} = 1 \\ (i i i)^{n} C_{r} =^{n} C_{n - r} \end{array}

Example 1:

\begin{array}{l} {Calculate the value of}^{7} C_{2} \\ ^{7} C_{2} = \frac{7!}{(7 - 2)! \times 2!} \\ = \frac{7!}{5! \times 2!} \\ = \frac{7 \times 6 \times 5!}{5! \times 2!} \\ = \frac{7 \times 6}{2 \times 1} \\ = 21 \end{array}

Calculator Computation:

Example 2:

There are 6 marbles, each with different colour, which are to be divided equally between 2 children. Find the number of different ways the division of the marbles can be done.

Solution:

Number of ways giving 3 marbles to the first child =

^{6} C_{3}

Number of ways giving the remaining 3 marbles =

^{3} C_{3}

So, the number of different ways the division of the marbles

\begin{array}{l} =^{6} C_{3} \times^{3} C_{3} \\ = 20 \times 1 \\ = 20 \end{array}

6.1 Permutations Part 2

Posted on April 23, 2020 by user

(C) Permutation of n Different Objects, Taken r at a Time

1. The number of permutations of n different objects, taken r at a time is given by:

2. A permutation of n different objects, taken r at a time, is an arrangement of a set of r objects chosen from n objects. The order of the objects in the chosen set is taken into consideration.

3. The number of permutations of n different objects, taken all at a time, is:

Example 1:

Evaluate each of the following.

\begin{array}{l} {(a)}^{5} P_{2} \\ {(b)}^{7} P_{3} \\ {(c)}^{9} P_{4} \end{array}

Solution:

\begin{array}{l} {(a)}^{5} P_{2} = \frac{5!}{(5 - 2)!} = \frac{5!}{3!} \\ = \frac{5 \times 4 \times 3!}{3!} = 5 \times 4 = 20 \\ {(b)}^{7} P_{3} = \frac{7!}{(7 - 3)!} = \frac{7!}{4!} \\ = \frac{7 \times 6 \times 5 \times 4!}{4!} = 7 \times 6 \times 5 = 210 \\ {(c)}^{9} P_{4} = \frac{9!}{(9 - 4)!} = \frac{9!}{5!} \\ = \frac{9 \times 8 \times 7 \times 6 \times 5!}{5!} = 9 \times 8 \times 7 \times 6 = 3024 \end{array}

Calculator Computation: